Question

I need help with this problem! I have posted a solution that got this review: You have translated the problem to show that someone can place a maximum of n(n-1) of figure 2 in the layer n. But you need to show that this formula is true! Do do this, color all triangles in a way such that every figure 2 covers one black and one white ruta- but that there only is n(n-1)/2 black rutorn in the layer n. PLEASE HELP!!

The first picture below consists of 60 small triangles and the other one of 2 same small triangles. What is the biggest amount of copies of the little picture that can be placed on the bigger picture?The pieces should be placed so the small triangles cover each other and 2 pieces not overlap. It is allowed to rotate the pieces. Note the hole in the middle of the large triangle PhotoCrid We will tep tep we Baric tik is ager tile So Awe hae no >for 2 we ha 1a tie withot urlpingSo, Fr 3 a hat 3 ily without ovrlaping- So, 3 hare we Cn- tles without n 2 (n-) So ovepng 2 Therufare Far8 Qayrsd would have maimum ot a fw withot 8728 tlas is 2 bnddayiaad Bu, mthe iven quustion thure is 2agyor tile A triangle hde which a tramgle that has Hence 27 tilas whot

Answer 1

The solution is correct as we can see that we can actually use 27 tiles and get a non-overlapping covering. But you have not acutally proved that it is the maximum as you are choosing a specific way of covering the tiles(you haven't mentioned it explicity but I think you are using pairing the triangles they come in the row while leaving the last triangle unpaired, for ex in case of the second layer, the first two triangles on the second while leaving the last one uncovered, in the rows where we have a gap a minor adjustment needs to be made, which has been done by you). There might be some other way of covering the bigger picture with smaller which might give space for more tiles. The review is simply asking you show that this can't be done The the review also gives you a now (assume that there is each row in the following fashion: the first triangle of each row is black and the colouring will be alternate, for ex, the second row will have the following colouring:black, white, black. In this colouring of the same colour share an edge with each other. Therefore when we place the small picture on the bigger one, its one triangle is bound to overlap with white triangle. So you can't have me any covering then the umber of white triangle, in other words, if W is the number of white triangles, and S is the number of copies of little picture that are allowed, we have S W Now our job is almost done, we just have to show that the nh layer wll have exactly n)white triangles. The nth row has 2n 1 triangles where only 1of these triangle is white. Summing k1= nn- Therefore we have S< n(n-1) arguments itself, we have S> n(n-1) Hence, S n(n-1) as covered row are Ones hint for the proof which I will explain hole for simplicity Colour the triangle in no observe that no two tile we then pieces of smaller picture in a nore 2 ws we have From your over all the ro W _ k-0 2 2 2 2